If one roots of k(x−1)2=8x−14,k>0,  is twice the other then the value of k is

Given equation is k(x−1)2=8x−14, k>0 It can be written as k(x2−2x+1)=8x−14 ⇒kx2−(2k+8)x+k+14=0 Let the roots be α  and 2α,  then sum of the roots : α+2α=  2k+8k And  product of the roots :   α(2α)  =  k+14k ⇒    α=  2k+83k      and 2α2=  k+14k ⇒    2(2k+83k)2=  k+14k      (∵ α=  2k+83k ) ⇒2(4k2+64+32k9k2)=k+14k ⇒    8k2+64k+128  =  9k2+126k ⇒    k2+62k−128=0 ⇒    (k+64)(k−2)  =0  ⇒  k=−64,  2. ∴ k=2       (∵k>0)